The mean and the standard deviation (S.D.) of five observations are 9 and 0, respectively. If one of the observation is increased such that the mean of the new set of five observations becomes 10, then their S.D. is:

We have five observations whose mean is given as $$\bar x = 9$$ and whose standard deviation is $$\sigma = 0$$.

First, recall the result: if the standard deviation of a data set is zero, all the observations are identical. Hence every one of the five numbers must be $$9$$. We may therefore write the original data set as $$\{9,\,9,\,9,\,9,\,9\}$$.

Because the mean equals the arithmetic average, the sum of these five numbers must be

$$\text{Original sum}=5\times 9=45.$$

Now one observation is increased (all other observations remain unchanged) so that the mean of the new data becomes $$10$$. Using the definition of mean once again, the sum of the new five observations must be

$$\text{New sum}=5\times 10=50.$$

The only change from the old sum $$45$$ to the new sum $$50$$ comes from increasing a single observation, so the amount of increase is

$$50-45 = 5.$$

Since the increased observation was originally $$9$$, its new value becomes

$$9+5=14.$$

Therefore, the new data set is

$$\{14,\,9,\,9,\,9,\,9\}.$$

Next, we compute the standard deviation of this new set. We shall use the population standard deviation formula

$$\sigma = \sqrt{\dfrac{\sum_{i=1}^{n}(x_i-\bar x)^2}{n}},$$

where $$n=5$$ and $$\bar x=10$$.

Compute each deviation from the mean:

$$ \begin{aligned} 14-10 &= 4,\\ 9-10 &= -1,\\ 9-10 &= -1,\\ 9-10 &= -1,\\ 9-10 &= -1. \end{aligned} $$

Square these deviations:

$$ \begin{aligned} 4^2 &= 16,\\ (-1)^2 &= 1,\\ (-1)^2 &= 1,\\ (-1)^2 &= 1,\\ (-1)^2 &= 1. \end{aligned} $$

Add the squared deviations:

$$16+1+1+1+1 = 20.$$

Divide by the number of observations $$n=5$$:

$$\dfrac{20}{5}=4.$$

Finally, take the square root to obtain the standard deviation:

$$\sigma = \sqrt{4}=2.$$

Hence, the correct answer is Option B.